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# Class Note for CSCI 124 at GW (8)

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This 4 page Class Notes was uploaded by an elite notetaker on Saturday February 7, 2015. The Class Notes belongs to a course at George Washington University taught by a professor in Fall. Since its upload, it has received 18 views.

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Date Created: 02/07/15

Linear Combinations and Spans Notes for CSCi 124224 Poorvi L Vora De nition A linear cambinatian ofn vectors vi 6 V is a vector ELI aivi for some ai 1 C F The values ai above are often referred to as the cae cients Example 1 2 1 2 75 2 2 5 3 7 7 3 72 3 4 2 12 is a linear combination of 2 1 2 2 7 3 7 3 3 4 2 A shorthand way of saying that a vector V is a linear combination of vi L1 is to say it lies in the span of these vectors The span is the set of all possible linear combinations De nition The span of a nite set of vectors B vi1 C V is 71 SpanB aivilai E F i1 How does one determine if a given vector V belongs to the span of vi 1 Further if it does how does one determine the values ai Answer solve a set of linear equations 1 Determining if v E Spanvz 121 1 The question is are there values a1 a2 an such that alvl a2V2 anvn V 2 Form a matrix A with vi1 as its columns A v1v2vn Vector SpacesVoraGWU 2 3 Write a vector 3 containing the unknown coef cients ai a1 02 an 4 Solve the set of linear equations Aa V using gaussian elimination 5 If there is exactly one solution or many V E SpanVi1 and the solutions provide the coef cients 6 If there are no solutions V is not in Spa1Vi 1 2 Examples Example 2 Determine if 2 V 72 3 belongs to SpanVl 1 where 1 1 1 V1 1 V2 7 1 V3 7 1 1 71 We would proceed as follows The question being asked is are there values a1 a2 and a3 such that 1 1 1 2 a1 1 a2 1 013 1 72 1 1 71 3 That is does the set of equations a1 a2 a3 2 a1 7 a2 a3 2 a1 a2 7 a3 3 have a solution We can solve this using gaussian elimination right away a1 a2 a3 2 7 202 74 7203 1 Vector SpacesVoraGWU 3 which gives us a2 2 a3 7 and a1 Which means a solution does exist and the given vector V may be expressed as a linear combination of the vectors Vi 91 We could also have written the above equation as a matrix equation Recall we did this when we were solving linear equations The matrix is constructed with the vectors Vl39 as columns v1 is the rst column of the matrix V2 is the second and so on 1 1 1 a1 2 1 71 1 a2 72 1 1 71 a3 3 The above set of equations can be solved by trying to invert the matrix if it is square or as above by using gaussian elimination which is usually the quickest and cleanest approach In the above example V E SpanB What happens when V dues nut belong to Spanvi1 Or when it does but there are many possible values of a1 a2 quotan Answer the same thing that happens when a set of equations has no solutions or has many solutions respectively Example 3 Determine if 2 72 3 belongs to Spanvi 1 where l l 0 V1 7 1 V2 7 1 V3 7 1 l l 0 1 1 0 a1 2 1 71 1 a2 72 1 1 0 a3 3 if it exists If you proceed with gaussian elimination you will be ne If you try to invert the matrix you won t be successful because its determinant is zero So always rst attempt gaussian elimination Don t forget to perform the same operations on the righthand side 1 1 0 a1 2 0 72 1 a2 74 0 0 0 a3 1 This gives you a nal equation of the form 0 1 This is an inconsistent set of equations which means that V is not in Spanvi 1 Vector SpacesVoraGWU 4 Example 4 Detennine if 3 4 3 is a linear combination of 1 1 0 V1 1 7V2 1 7V3 1 1 1 1 0 1 1 0 a1 3 1 71 1 a2 4 1 1 0 a3 3 if it exists Gaussian elimination gives 1 1 0 a1 3 0 72 1 a2 1 0 0 0 a3 0 where the nal equation is of the form 0 O This means you have a consistent set of equations with several solutions One solution is 13 1 a2 0 a1 3 That is 31071114 2

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